Countable intersection of F-sigma sets

[jbunniii]

My question concerns F_\sigma subsets of \mathbb{R}. An F_\sigma set is one which can be expressed as a countable union of closed sets.

I have several books that state that a countable intersection of F_\sigma sets need not be an F_\sigma set (indeed, such sets have their own designation, F_{\sigma\delta}), but none of them gives a counterexample. Does anyone know one offhand? This isn't homework/coursework; I'm just curious and haven't been able to come up with one. I would prefer a concrete example if possible, not just an existence proof.

Relevant facts: Clearly a countable UNION of F_\sigma sets is F_\sigma. All open and closed sets in \mathbb{R} are F_\sigma. \mathbb{Q} is an F_\sigma set as it is a countable union of singletons. The only concrete example of a set I know is not F_\sigma is \mathbb{R}\setminus\mathbb{Q}, the set of irrationals.

 

[micromass]

\mathbb{R}\setminus \mathbb{Q} = \bigcap_{q\in \mathbb{Q}} \mathbb{R}\setminus \{q\}

 

[jbunniii]

\mathbb{R}\setminus \mathbb{Q} = \bigcap_{q\in \mathbb{Q}} \mathbb{R}\setminus \{q\}

Yes, I just had that realization while thinking about it again. I had been trying some more convoluted way of finding a sequence of open sets that decreased to \mathbb{R}\setminus\mathbb{Q}, but missed the most obvious one!

 

[jbunniii]

The above construction also shows that \mathbb{R}\setminus\mathbb{Q} is a G_\delta set (i.e. a countable intersection of open sets), which also follows immediately from the fact that its complement \mathbb{Q} is a countable union of singletons, hence a F_\sigma set.

So \mathbb{R}\setminus\mathbb{Q} is in some sense an atypical member of F_{\sigma\delta} because it is also in G_\delta. Similarly, \mathbb{Q} is an atypical member of F_{\sigma\delta} because it is also in F_\sigma.

This led me to look for an example of a "pure" F_{\sigma\delta} set, one which is in neither F_\sigma nor G_\delta.

I think that S = (\mathbb{Q} \cap (-\infty,0)) \cup ((\mathbb{R}\setminus\mathbb{Q}) \cap (0,\infty)) is such a set. (i.e. the union of all negative rationals and all positive irrationals.) Here is my reasoning:

If S were a F_\sigma set, then S \cap (0,\infty) would also be F_\sigma, but that is not the case, because S \cap (0,\infty) = ((\mathbb{R}\setminus\mathbb{Q}) \cap (0,\infty)). Similarly, S is not a G_\delta set. However, I can form a descending sequence of F_\sigma sets whose intersection is S by starting with (Q \cap(-\infty,0)) \cup (0,\infty) and removing one positive rational at a time.

Then I started thinking about what a "pure" F_{\sigma\delta\sigma} set would look like, and it made me tired.